Evaluate the integral #int \ 1/(x^2sqrt(x^2-9)) \ dx #?

3 Answers
Oct 18, 2017

# sqrt(x^2-9)/(9|x|)+C.#

Explanation:

We substitute, #x=3secy rArr dx=3secytanydy.#

#:. I=int1/(x^2sqrt(x^2-9))dx,#

#=int{(3secy tany)dy}/{9sec^2ysqrt(9sec^2y-9),#

#=int{(cancel(3)secycancel(tany))dy}/{9sec^2y*cancel(3tany)},#

#=1/9int1/secydy=1/9intcosydy,#

#=1/9siny,#

#=1/9sqrt(1-cos^2y),#

#=1/9sqrt(1-1/sec^2y),#

#=1/9sqrt(1-(3/x)^2).......................[because, x=3secy],#

#=1/9sqrt(x^2-9)/|x|.#

# rArr I=sqrt(x^2-9)/(9|x|)+C.#

Enjoy Maths.!

Oct 18, 2017

# int \ 1/(x^2sqrt(x^2-9)) \ dx = 1/9 sqrt(1-9/x^2) + C #

Explanation:

We seek:

# I = int \ 1/(x^2sqrt(x^2-9)) \ dx #

We can write the integral as follows:

# I = int \ 1/(x^2sqrt(9(x^2/9-1))) \ dx #
# \ \ = int \ 1/(x^2sqrt(9)sqrt((x/3)^2-1)) \ dx #
# \ \ = 1/3 \ int \ 1/(x^2 sqrt((x/3)^2-1)) \ dx #

Let us attempt a substitution of the form:

# 3sec theta = x => 3sectheta tantheta (d theta)/dx = 1 #

Then substituting into the integral, we get:

# I = 1/3 \ int \ 1/((3sec theta)^2 sqrt((sec theta)^2-1)) \ 3sectheta tantheta \ d theta #

# \ \ = int \ (sectheta tantheta)/(9sec^2 theta sqrt(sec^2theta-1)) \ d theta #

# \ \ = 1/9 \ int \ (tantheta)/(sec theta sqrt(tan^2 theta)) \ d theta #

# \ \ = 1/9 \ int \ (tantheta)/(sec theta tan theta) \ d theta #

# \ \ = 1/9 \ int \ (1)/(sec theta) \ d theta #

# \ \ = 1/9 \ int \ cos theta \ d theta #

# \ \ = 1/9 sin theta + C #

# \ \ = 1/9 sqrt(1-cos^2 theta) + C #

# \ \ = 1/9 sqrt(1-(1/sec^2 theta)) + C #

And we can now restore the earlier substitution:

# I = 1/9 sqrt(1-(1/(x/3)^2)) + C #
# \ \ = 1/9 sqrt(1-(1/(x^2/9)) + C #
# \ \ = 1/9 sqrt(1-9/x^2) + C #

Oct 18, 2017

#int dx/(x^2sqrt(x^2-9)) = sqrt(x^2-9)/(9x)+ C#

Explanation:

Restricting to the interval #x in (3,+oo)#, substitute:

#x=3sect#

#dx = 3 sect tant dt#

with #t in (0,pi/2)# and get:

#int dx/(x^2sqrt(x^2-9)) = 3 int (sect tant dt)/(9sec^2t sqrt(9sec^2t -9))#

#int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect sqrt(sec^2t -1))#

use now the trigonometric identity:

#sec^2t -1 = 1/cos^2-1 = (1-cos^2t)/cos^2t =sin^2t/cos^2t = tan^2t#

so that:

#int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect sqrt(tan^2t))#

and as in the selected interval the tangent is positive:

#int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect tant) #

#int dx/(x^2sqrt(x^2-9)) = 1/9 int dt/sect #

#int dx/(x^2sqrt(x^2-9)) = 1/9 int costdt#

#int dx/(x^2sqrt(x^2-9)) = 1/9 sint + C#

To undo the substitution note that:

#x= 3sect = 3/cost = 3/sqrt(1-sin^2t)#

#sqrt(1-sin^2t) = 3/x#

#1-sin^2t = 9/x^2#

#sin^2t = 1-9/x^2#

and as in the interval the sine is positive:

#sint = sqrt(x^2-9)/x#

So:

#int dx/(x^2sqrt(x^2-9)) = sqrt(x^2-9)/(9x)+ C#

And using the substitution #x=-sect# for #x in (-oo,-3)# we obtain the same result.